Question

数列的前十项中有多少个质数？ $121, 11211, 1112111, \ldots$

How many of the first ten numbers of the sequence $121, 11211, 1112111, \ldots$ are prime numbers?

$0$

$1$

$2$

$3$

$4$

答案：A

知识点：质数因式分解

难度评级：1140

解答：

我们断言这些数都不可能是质数。

第 $n$ 项可写成这说明该数可写成两个大于 $1$ 的整数的乘积，所以没有质数。 $\sum_{k=0}^{2n} 10^k + 10^n = \sum_{k=0}^{n} 10^k +$ $\sum_{k=n}^{2n} 10^k =\sum_{k=0}^{n} 10^k + \sum_{k=0}^{n} 10^k \cdot 10^n$ $= (10^n+1)(\sum_{k=0}^{n} 10^k).$

所以正确答案是 A。

We claim that none of these numbers can ever be prime.

We prove this claim by noticing that the $n$ th number is $\sum_{k=0}^{2n} 10^k + 10^n = \sum_{k=0}^{n} 10^k +$ $\sum_{k=n}^{2n} 10^k =\sum_{k=0}^{n} 10^k + \sum_{k=0}^{n} 10^k \cdot 10^n$ $= (10^n+1)(\sum_{k=0}^{n} 10^k).$ This shows that the number can be written as the product of two numbers greater than $1,$ so there are no primes.

Thus, the answer is A .

2022 AMC 10B 第 6 题

答案：A

难度评级：1140

解答：

其他年份的第 6 题